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Proof step 82 from proof inv-get-inverse-succ (previous step)

# nσ = ∞, stutf(nσ), nσ[n1] ≤ nσ[m], ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ nσ[n1 + 1] = nσ[n1] + 1

User applied rule all left

And obtained the following new goals

Proof step 83:
# nσ = ∞, stutf(nσ), nσ[n1] ≤ nσ[m], ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, m < m + 1 → nσ[m + 1] = nσ[m] + 1 ⊦ nσ[n1 + 1] = nσ[n1] + 1

Rule argument

pos: left 5 terms: m + 1 abort: F discard: T computed: T

Additional information

Used terms: m + 1